the equation is

$\ln(x+2)=\ln e^{\ln2} - \ln x$

How do I solve for $x$?

  • 1
    $\begingroup$ $e^{\ln 2} = 2$, for starters $\endgroup$ – The Chaz 2.0 Oct 23 '13 at 4:03
  • $\begingroup$ can you explain why ? $\endgroup$ – user293849 Oct 23 '13 at 4:26
  • $\begingroup$ @user293849: Think about it: What does "$\ln 2$" mean? $\endgroup$ – Blue Oct 23 '13 at 4:27
  • $\begingroup$ @user293849 Look up the logarithm identities $\ln a^b = b \ln a$. In your case $b=\ln 2$ and $\ln e=1$. Please have a look here en.wikipedia.org/wiki/List_of_logarithmic_identities $\endgroup$ – triomphe Oct 23 '13 at 4:32

Your equation is equivalent to $$\ln (x+2)+\ln(x)=\ln2.$$ Then use logarithm identity $\ln ab=\ln a +\ln b$\$ $$\ln (x+2)x=\ln2.$$ Now take inverse of both sides $$(x+2)x=2.$$ Now you can solve the quadratic equation and select the appropriate $x$ value. You get $x=-1\pm\sqrt 3.$ Note that you can not take $x<0$ as $\ln x$ is not defined for negative $x$. So your answer is $x=-1+\sqrt 3$ which is approximately equal to $0.732$.


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